# Question f4042

May 2, 2017

#### Answer:

The percent yield is 87.3 %.

#### Explanation:

This is a limiting reactant problem.

We know that we will need a balanced equation, masses, and moles, so let's gather all the information in one place.

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m m m} 170.48 \textcolor{w h i t e}{m m m} 380.12 \textcolor{w h i t e}{m m m l l} 434.60$
$\textcolor{w h i t e}{m m m m m m m} \text{3CuCl"_2 + color(white)(ll)"2Na"_3"PO"_4 → "Cu"_3("PO"_4)_2 + "6NaCl}$
$\text{Mass/g:} \textcolor{w h i t e}{m m m l} 1.780 \textcolor{w h i t e}{m m m l l} 1.773$
$\text{Moles:"color(white)(mmm)"0.010 441"color(white)(mll)"0.004 664}$
$\text{Divide by:} \textcolor{w h i t e}{m m m l} 3 \textcolor{w h i t e}{m m m m m l l} 2$
$\text{Moles rxn:"color(white)(m)"0.003 480"color(white)(mll)"0.002 332}$

${\text{Moles of CuCl"_2 = 1.780 color(red)(cancel(color(black)("g CuCl"_2))) × "1 mol CuCl"_2/(170.48 color(red)(cancel(color(black)("g CuCl"_2)))) = "0.010 441 mol CuCl}}_{2}$

${\text{Moles of Na"_3"PO"_4 = 1.773 color(red)(cancel(color(black)("g Na"_3"PO"_4))) × ("1 mol Na"_3"PO"_4)/(380.12 color(red)(cancel(color(black)("g Na"_3"PO"_4)))) = "0.004 664 mol Na"_3"PO}}_{4}$

Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will produce.

You simply divide the number of moles by the corresponding coefficient in the balanced equation.

I did that for you in the chart above.

$\text{Na"_3"PO"_4}$ is the limiting reactant because it gives the fewest "moles of reaction".

Calculate the theoretical yield

"Theor. yield" = "0.004 664" color(red)(cancel(color(black)("mol Na"_3"PO"_4))) × (1 color(red)(cancel(color(black)("mol Cu"_3("PO"_4)_2))))/(2 color(red)(cancel(color(black)("mol Na"_3"PO"_4)))) × ("434.60 g Cu"_3("PO"_4)_2)/(1 color(red)(cancel(color(black)("mol Cu"_3("PO"_4)_2)))) = "1.0136 g Cu"_3("PO"_4)_2

Calculate the percentage yield

"% yield" = "actual"/"theoretical" × 100 % = (0.885 color(red)(cancel(color(black)("g"))))/(1.0136 color(red)(cancel(color(black)("g")))) × 100% = 87.3 %#